What does a cell use to eliminate a substance that is too large to leave by diffusion?

I want to take a picture of the question and get the answer

OleMash [197]

When you take a picture it pulls up people who asked the same question and they’ll have a answer you can use

5 0

2 years ago

The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to

DedPeter [7]

Answer:

$9.4\cdot 10^{10} m$

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

$\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}$

where

$r_o$ is the distance of the new object from the sun (orbital radius)

$T_o=180 d$ is the orbital period of the object

$r_e = 1.50\cdot 10^{11} m$ is the orbital radius of the Earth

$T_e=365 d$ is the orbital period the Earth

Solving the equation for $r_o$ , we find

$r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m$

3 0

3 years ago

A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 88.4 Hz, determine the first three

yulyashka [42]

Answer with Explanation:

We are given that

Fundamental frequency,f=88.4 Hz

Speed of sound,v=343 m/s

We have to find the first three overtones.

Tube is closed

The overtone of closed pipe is equal to odd number of fundamental frequency.

Therefore, the overtone of tuba

$f'=nf$

Where n=3,5,7,..

Substitute n=3

$f'=3\times 88.4=265.2Hz$

For second overtone

$f'=5\times 88.4=442Hz$

For third overtone

$f'=7\times 88.4=618.8Hz$

4 0

3 years ago

Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i

densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0

3 years ago

A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to

Len [333]

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

5 0

3 years ago